question_answer 3)

\[PC{{l}_{5}},PC{{l}_{3}}\]and\[C{{l}_{2}}\] are at equilibrium at\[500K\]and having concentration 1.59 M\[PC{{l}_{3}},1.59M\,C{{l}_{2}}\] and\[1.41\,M\,PC{{l}_{5}}\]. Calculate \[{{K}_{c}}\] for the reaction:

Answer:

\[{{K}_{c}}\]for the reaction will be: \[{{K}_{c}}=\frac{[PC{{l}_{5}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{1.59\times 1.59}{1.41}\] \[=1.79M\]